如何确保类的每个方法都首先调用其他方法？ [英] How to ensure that every method of a class calls some other method first?

问题描述

我有：

class Foo {
   public:
      void log() { }

      void a() {
         log();
      }

      void b() {
         log();
      }
};

有没有一种方法可以让每种方法 Foo ，调用 log（），但无需我显式键入log（）作为每个函数的第一行？
我想这样做，这样我就可以向每个函数添加行为，而不必遍历每个函数并确保进行了调用，并且还可以在添加新函数时自动添加代码。 ..

Is there a way that I can have each method of Foo, call log(), but without me having to explicitly type log() as the first line of each function ? I want to do this, so that I can add behaviour to each function without having to go through each function and make sure the call is made, and also so that when I add new functions, the code is automatically added...

这甚至可能吗？我无法想象如何使用宏执行此操作，因此不确定从哪里开始...到目前为止，我唯一想到的方法是添加预构建步骤，以便在编译之前扫描文件并编辑源代码，但这似乎不是很聪明。...

Is this even possible ? I can't imagine how to do this with macro's, so not sure where to begin... The only way I have thought of so far, is to add a "pre-build step", so that before compiling I scan the file and edit the source code, but that doesn't seem very intelligent....

编辑：只是为了澄清-我不希望log（）自己调用明显。它不需要成为课程的一部分。

Just to clarify - I don't want log() to call itself obviously. It doesn't need to be part of the class.

编辑：我更喜欢使用跨平台工作的方法，并且仅使用stl。

I would prefer using methods that would work cross platform, and using only the stl.

推荐答案

由于 operator-> 的不寻常属性，我们可以在任何成员访问之前注入代码，但需要付出稍微弯曲的语法：

Thanks to the unusual properties of operator ->, we can inject code before any member access, at the expense of a slightly bent syntax:

// Nothing special in Foo
struct Foo {
    void a() { }
    void b() { }
    void c() { }
};

struct LoggingFoo : private Foo {
    void log() const { }

    // Here comes the trick
    Foo const *operator -> () const { log(); return this; }
    Foo       *operator -> ()       { log(); return this; }
};

用法如下：

LoggingFoo f;
f->a();

在Coliru上实时观看

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